Set of Finite Subsets of Countable Set is Countable/Proof 3

Theorem

Let $A$ be a countable set.

Then the set of finite subsets of $A$ is countable.

Let $A^{\paren n}$ be the set of subsets of $A$ with no more than $n$ elements.

Thus:

$A^{\paren 0} = \set \O$

$A^{\paren 1} = A^{\paren 0} \cup \set {\set a: a \in A}$

and $\forall n \ge 0$:

$A^{\paren {n + 1} } = \set {a^{\paren n} \cup a^{\paren 1}: a^{\paren n} \in A^{\paren n} \land a^{\paren 1} \in A^{\paren 1} }$

Let us verify by induction that each $A^{\paren n}$ is countable.

Let $A$ be countable.

$A^{\paren 1}$ is countable, as its cardinality is $1 + \card A$.

Suppose $A^{\paren n}$ is countable.

Then by Union of Countable Sets of Sets, so $A^{\paren {n + 1} }$ also countable.

By induction, each $A^{\paren n}$ is countable.

Denote with $A^f$ the set of finite subsets of $A$.

It is apparent that every finite subset is in some $A^{\paren n}$, and so:

$A^f = \ds \bigcup_{n \mathop \in \N} A^{\paren n}$

$\blacksquare$

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.

1964: Steven A. Gaal: Point Set Topology ... (previous) ... (next): Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability